Question Number 36818 by maxmathsup by imad last updated on 06/Jun/18
$${find}\:{f}\left({a}\right)\:=\:\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{ax}^{\mathrm{2}} }}\:\:{with}\:{a}\:{from}\:{R}\:. \\ $$
Commented by prof Abdo imad last updated on 06/Jun/18
$${case}\:\mathrm{1}\:\:{a}>\mathrm{0}\:\:{changement}\:\sqrt{{a}}{x}=\:{sint}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int\:\:\:\:\:\frac{\mathrm{1}}{{cost}}\:\frac{{cost}\:{dt}}{\:\sqrt{{a}}}\:=\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\:{arcsin}\left(\sqrt{{a}}\:{x}\right)\:+{c} \\ $$$${case}\:\mathrm{2}\:\:{a}<\mathrm{0}\:\:{changement}\:\sqrt{−{a}}{x}={sh}\left({t}\right)\:{give} \\ $$$${f}\left({a}\right)\:=\:\int\:\:\:\frac{\mathrm{1}}{{ch}\left({t}\right)}\:\frac{{ch}\left({t}\right)}{\:\sqrt{−{a}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{−{a}}}\:{argsh}\left(\sqrt{−{a}}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{−{a}}}\:{ln}\left\{\:{x}\sqrt{−{a}}\:\:+\sqrt{\mathrm{1}−{ax}^{\mathrm{2}} }\:\:\right)\:+{c}\:. \\ $$
Answered by MJS last updated on 06/Jun/18
![∫(dx/( (√(1−ax^2 ))))= [t=(√a)x → dx=(dt/( (√a)))] =(1/( (√a)))∫(dt/( (√(1−t^2 ))))=(1/( (√a)))arcsin t= ((√a)/a)arcsin (√a)x +C { ((a>0 (1/( (√a)))arcsin (√a)x +C)),((a=0 x+C)),((a<0 (1/( (√(−a))))arcsinh (√(−a))x +C)) :}](https://www.tinkutara.com/question/Q36833.png)
$$\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{ax}^{\mathrm{2}} }}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{{a}}{x}\:\rightarrow\:{dx}=\frac{{dt}}{\:\sqrt{{a}}}\right] \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{{a}}}\mathrm{arcsin}\:{t}= \\ $$$$\frac{\sqrt{{a}}}{{a}}\mathrm{arcsin}\:\sqrt{{a}}{x}\:+{C} \\ $$$$\begin{cases}{{a}>\mathrm{0}\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{{a}}}\mathrm{arcsin}\:\sqrt{{a}}{x}\:+{C}}\\{{a}=\mathrm{0}\:\:\:\:\:{x}+{C}}\\{{a}<\mathrm{0}\:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{−{a}}}\mathrm{arcsinh}\:\sqrt{−{a}}{x}\:+{C}}\end{cases} \\ $$